Evaluate :
`int(log|sin x|)/(tan x)dx`

Answer

To evaluate the integral \( \int \frac{\log |\sin x|}{\tan x} \, dx \), we can follow these steps:

### Step 1: Substitution
Let \( t = \log |\sin x| \). Then, we need to find \( dt \) in terms of \( dx \).

### Step 2: Differentiate \( t \)
Differentiating \( t \) with respect to \( x \):
\[ dt = \frac{1}{\sin x} \cdot \cos x \, dx = \frac{\cos x}{\sin x} \, dx = \cot x \, dx \]
This implies:
\[ dx = \frac{dt}{\cot x} = \frac{dt}{\frac{\cos x}{\sin x}} = \frac{\sin x}{\cos x} \, dt = \tan x \, dt \]

### Step 3: Rewrite the Integral
Now substitute \( t \) and \( dx \) into the integral:
\[ \int \frac{\log |\sin x|}{\tan x} \, dx = \int \frac{t}{\tan x} \cdot \tan x \, dt = \int t \, dt \]

### Step 4: Integrate
Now, we can integrate:
\[ \int t \, dt = \frac{t^2}{2} + C \]

### Step 5: Substitute Back
Now substitute back \( t = \log |\sin x| \):
\[ \frac{(\log |\sin x|)^2}{2} + C \]

### Final Answer
Thus, the final answer is:
\[ \int \frac{\log |\sin x|}{\tan x} \, dx = \frac{(\log |\sin x|)^2}{2} + C \]

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